# How many moles of hydrogen gas can be evolved from the action of excess hydrochloric acid on a mass of 4.8*g of magnesium metal?

Apr 6, 2017

Approx. $4.8 \cdot L$

#### Explanation:

We assess the following chemical reaction:

$M g \left(s\right) + 2 {H}_{3} {O}^{+} \rightarrow M {g}^{2 +} + {H}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right)$

And thus $\text{moles of metal}$ $\equiv$ $\text{moles of dihydrogen gas.}$

$\text{Moles of metal}$ $=$ $\text{Mass"/"Molar mass}$

$= \frac{4.8 \cdot g}{24.3 \cdot g \cdot m o {l}^{-} 1} = 0.198 \cdot m o l$.

And this $0.198 \cdot m o l$ $\text{dihydrogen gas is evolved}$.

The molar volume of an ideal gas under standard conditions is $24 \cdot {\mathrm{dm}}^{3}$. (The definitions now vary across the different syllabuses, so I cannot be much help to you here). And so the volume of gas evolved is:

0.198*molxx24*dm^3*mol^-1=??dm^3.